The Algebraic Closure of a Galois Field
In the present chapter, we study the algebraic closure of a Galois field. For this purpose, we also provide some basic results on algebraic extensions, extending the material covered in Chapterss 2 and 3 .
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The Algebraic Closure of a Galois Field
Abstract In the present chapter, we study the algebraic closure of a Galois field. For this purpose, we also provide some basic results on algebraic extensions, extending the material covered in Chapters 2 and 3.
4.1 Preliminaries on Algebraic Extensions In this section, we collect some fundamental results on algebraic extensions. Let us first recall some basic definitions from Section 2.4. Given a field extension E/F and an element v ∈ E, the evaluation homomorphism at v is defined by Γv : F[x] → E, f (x) → f (v). Recall that v is said to be algebraic over F provided that Γv is not injective; in this case, the unique monic generator mpolv (x) of kerΓv is called the minimal polynomial of v over F. In what follows, the minimal polynomial will also be denoted by mpolv,F (x), if we wish to emphasize the underlying field. The extension E/F is algebraic if every v ∈ E is algebraic over F. We also say that a field extension E/F is finite if the degree [E : F] is finite. Lemma 4.1.1. Every field extension E/F with finite degree is algebraic. Conversely, if E/F is an algebraic extension and S ⊆ E is finite, then F(S)/F is finite. Proof. The first assertion is clear, as the intermediate field F(v) of E/F generated by v ∈ E is isomorphic to F[x]/ kerΓv . Since E/F is finite and [F(v) : F] ≤ [E : F], the kernel of Γv has to be non-trivial. The partial converse in the second assertion obviously holds if S = {u} is a singleton. We proceed using induction on the cardinality of S. Choose some u ∈ S, put S := S \ {u}, and let K = F(u) and L = K(S ). Since E/K is algebraic, L/K is a finite extension by induction. As K/F is a finite extension, Lemma 3.1.1 shows that © Springer Nature Switzerland AG 2020 D. Hachenberger and D. Jungnickel, Topics in Galois Fields, Algorithms and Computation in Mathematics 29, https://doi.org/10.1007/978-3-030-60806-4_4
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4 The Algebraic Closure of a Galois Field
L/F is likewise finite. Since F(S) is an intermediate field of L/F, we conclude that F(S) is a finite extension over F. Proposition 4.1.2. Consider a field extension E/F, and let K be an intermediate field. Then E/F is algebraic if and only if both E/K and K/F are algebraic. Proof. It is clear that E/F can only be algebraic if both E/K and K/F are algebraic. For the converse, let u be an arbitrary element of E; by hypothesis, u is algebraic over K. Let S ⊆ K be the set of coefficients of mpolu,K (x), and consider the subfield L := F(S) of K. By Lemma 4.1.1, L/F is a finite extension. Moreover, [L(u) : L] = [K(u) : K], since mpolu,K (x) = mpolu,L (x) (by the definition of L); now Lemma 3.1.1 shows that the extension L(u)/F is finite. Since F(u) is an intermediate field of L(u)/F, we obtain that F(u)/F is also finite, so that u is algebraic over F. Corollary 4.1.3. Consider a field extension E/F and assume that u, v ∈ E such that u is algebraic over F and v is algebraic over F(u). Then v is also algebraic over F. Proof. The extension F(u, v)/F is finite, as F(u, v)/F(u) and F(u)/F are finite. Theorem 4.1.4. Let
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